I'll try and clarify a few of my earlier points.  Bill has pointed out
that the *very* recent ones, on which I hope you don't have any duds to
experiment, are a "minimalist" design with the electronics (1 chip) back
on the motherboard, leaving only three wires to the coils and maybe two
for the hall sensor.

  Note that the single hall sensor on the periphery is for the "index"
to indicate sector zero and plays no part in the motor operation.

Russell McMahon wrote:

> The centre is the common of all coils does not need to be connected to
> anything else (but can be if desired - see below).

  The winding will be a "star", not a "delta" (triangle).  This makes
better use of the turns.  The common ("centre") of the star is one end
of three windings, which are then in series with their opposites, either
in phase if the number of pole *pairs* is even (see below) or opposing
phase if the number is odd.

  Again, best use of the windings is made by using a (3-leg) "H" bridge
or "bipolar" driver rather than N-channel drivers and returning the
centre tap to supply.

> For mechanical reasons there are liable to be X North-South pairs of
> permanent magnets on the rotor where X is NOT a multiple of 3.

  Yes, if it were, all three coil pairs would attract at once.  But it's
at least two.  Each pole is about the size of one coil.  Estimate from
there.  Alternatively hold one pole of a magnet next to edge of the
rotor and count how many "cogs" you feel in one revolution = the number
of pole pairs.

> Because all coils are commoned you must apply voltage to 2 coils at
> once to cause current to flow.  When you do this one coil will produce
> a N pole and the other a S (as current goes into one and out of the
> other using consistent descriptions of in and out.  After applying
> power to 2 leads for a certain time the motor will have rotated to its
> stable position (one rotor N will have been attracted to and aligned
> with a stator S (or vice versa) and the other rotor pole will be
> repelled and now be mid-way between two same sign stator poles).

  Well, not actually a "stable" position as this is not supposed to act
as a stepper!

> Power must now be applied to one of the previously activated leads
> (but with opposite polarity) and to the previously unactivated lead.

  Not quite.  The previously activated lead retains the same polarity,
the opposite polarity goes to the previously idle lead.  Thus:

Full stepping 1 coil:
   1} 0 + + 0 - - 0 + + 0 - - 0 + + 0 - - 0
   2} - - 0 + + 0 - - 0 + + 0 - - 0 + + 0 -
   3} + 0 - - 0 + + 0 - - 0 + + 0 - - 0 + +

Full stepping 2 coil (not recommended):
   1} + + + - - - + + + - - - + + + - - - +
   2} - - + + + - - - + + + - - - + + + - -
   3} + - - - + + + - - - + + + - - - + + +

Half stepping:
   1} + + + + 0 - - - - - 0 + + + + + 0 - - - - - 0 + + + + + 0 - - - -
   2} - - 0 + + + + + 0 - - - - - 0 + + + + + 0 - - - - - 0 + + + + + 0
   3} 0 - - - - - 0 + + + + + 0 - - - - - 0 + + + + + 0 - - - - - 0 + +

> This process continues indefinitely and the rotor rotates :-).

> Position may be detected by 2 sensors (almost always hall effect
> sensors) which determine when the stator inputs should be changed by
> determining rotor alignment.

  You'd need three sensors = 3 phases.

 (Synchronous drive)
> Using sinusoidal voltages rather than square waves will assist this
> process.

  Why?  Using three phases (as in mains power transition) means there
are no "dead spots" in torque.

> If you must drive it without original drivers and power doesn't have
> to be maximised then you may be able to drive it "half wave".  This is
> very easy using square waves.  The centre is now connected to eg V+
> and up to 2 leads are taken low at any time via driver transistors,
> FETs or whatever.  Lends itself nicely to PIC control and no high side
> drivers needed.

  But you lose 50% efficiency.  Corresponds to unipolar drive of
steppers, a previously visited topic.

  Hope this makes things clearer.  Probably still forgot to answer a few
points though...
--
  Cheers,
        Paul B.